Find the best way to solve introduction to trigonometry question of ncert class 10 . What is the easy way to solve the exercise 8.1 question no. 3 , this question is tricky please help me to find the simplest way . If sin A = 3/4, Calculate cos A and tan A.

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Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC

^{2}=AB^{2 }+ BC^{2}Substitute the value of AC and BC

(4k)

^{2}=AB^{2}+ (3k)^{2}16k

^{2}−9k^{2 }=AB^{2}AB

^{2}=7k^{2}Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

Therefore, tan A = 3/√7